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题目1 ZJOI 最小割
题目大意：
求一个无向带权图两点间的最小割，询问小于等于c的点对有多少。
算法讨论： 最小割 分治
代码:
#include <cstdlib>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <vector>
using namespace std;
const int N = 150 + 5;
const int M = 3000 + 5;
const int oo = 0x3f3f3f3f;
#define inf oo
int n, m;
bool mark[N];
int a[N], tmp[N], ans[N][N];
struct Edge {
int from, to, cap, flow;
Edge(int u = 0, int v = 0, int c = 0, int f = 0) :
from(u), to(v), cap(c), flow(f) {}
};
struct Dinic {
int n, mm, s, t;
int dis[N], cur[N], que[N * 10];
bool vis[N];
vector <Edge> edges;
vector <int> G[N];
void clear() {
for(int i = 0; i <= n; ++ i) G[i].clear();
edges.clear();
}
void add(int from, int to, int cap) {
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, cap, 0));
mm = edges.size();
G[from].push_back(mm - 2);
G[to].push_back(mm - 1);
}
bool bfs() {
int head = 1, tail = 1;
memset(vis, false, (n + 1) * sizeof (bool));
dis[s] = 0; vis[s] = true; que[head] = s;
while(head <= tail) {
int x = que[head];
for(int i = 0; i < (signed) G[x].size(); ++ i) {
Edge &e = edges[G[x][i]];
if(!vis[e.to] && e.cap > e.flow) {
vis[e.to] = true;
dis[e.to] = dis[x] + 1;
que[++ tail] = e.to;
}
}
++ head;
}
return vis[t];
}
int dfs(int x, int a) {
if(x == t || a == 0) return a;
int flw = 0, f;
for(int &i = cur[x]; i < (signed) G[x].size(); ++ i) {
Edge &e = edges[G[x][i]];
if(dis[e.to] == dis[x] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
e.flow += f; edges[G[x][i] ^ 1].flow -= f; flw += f; a -= f;
if(a == 0) break;
}
}
return flw;
}
int maxflow(int s, int t) {
this->s = s; this->t = t;
int flw = 0;
while(bfs()) {
memset(cur, 0, sizeof cur);
flw += dfs(s, oo);
}
return flw;
}
void rebuild() {
for(int i = 0; i < (signed) edges.size(); ++ i)
edges[i].flow = 0;
}
void dfs(int u) {
mark[u] = true;
for(int i = 0; i < (signed) G[u].size(); ++ i) {
Edge e = edges[G[u][i]];
if(!mark[e.to] && e.cap > e.flow) {
dfs(e.to);
}
}
}
}net;
void Divide(int l, int r) {
if(l >= r) return;
net.rebuild();
int nowflow = net.maxflow(a[l], a[r]);
memset(mark, false, (n + 1) * sizeof (bool));
net.dfs(a[l]);
for(int i = 1; i <= n; ++ i)
if(mark[i])
for(int j = 1; j <= n; ++ j)
if(!mark[j])
ans[i][j] = min(ans[i][j], nowflow), ans[j][i] = ans[i][j];
int L = l, R = r;
for(int i = l; i <= r; ++ i)
if(mark[a[i]]) tmp[L ++] = a[i];
else tmp[R --] = a[i];
for(int i = l; i <= r; ++ i) a[i] = tmp[i];
Divide(l, L - 1); Divide(R + 1, r);//这不能二分一个Mid，因为S集和T集的大小不一定相同
}
int main() {
int T, u, v, c, q;
scanf("%d", &T);
while(T --) {
net.clear();
scanf("%d%d", &n, &m); net.n = n;
for(int i = 1; i <= n; ++ i) a[i] = i;
for(int i = 1; i <= n; ++ i)
for(int j = 1; j <= n; ++ j) ans[i][j] = inf;
for(int i = 1; i <= m; ++ i) {
scanf("%d%d%d", &u, &v, &c);
net.add(u, v, c);
}
Divide(1, n);
scanf("%d", &q);
for(int i = 1; i <= q; ++ i) {
int tmp = 0;
scanf("%d", &c);
for(int u = 1; u <= n; ++ u) {
for(int v = u + 1; v <= n; ++ v) {
if(ans[u][v] <= c) {
++ tmp;
}
}
}
printf("%d\n", tmp);
}
puts("");
}
return 0;
}
题目2 CQOI2016 不同的最小割
题目大意：
求所有点对问不同的最小割数目。
算法讨论： 最小割 分治
和上面的一个题有区别么？
代码：
#include <cstdlib>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <set>
using namespace std;
const int N = 850 + 5;
const int M = 8500 + 5;
const int oo = 0x3f3f3f3f;
int n, m;
int ans[N][N], tmp[N], a[N];
bool mark[N];
set <int> lts;
struct Edge {
int from, to, cap, flow;
Edge(int u = 0, int v = 0, int cap = 0, int flow = 0):
from(u), to(v), cap(cap), flow(flow) {}
};
struct Dinic {
int n, m, s, t;
int dis[N], cur[N], que[N * 10];
bool vis[N];
vector <Edge> edges;
vector <int> G[N];
void add(int from, int to, int cap) {
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, cap, 0));
m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
bool bfs() {
int head = 1, tail = 1;
memset(vis, false, (n + 1) * sizeof (bool));
dis[s] = 0; vis[s] = true; que[head] = s;
while(head <= tail) {
int x = que[head];
for(int i = 0; i < (signed) G[x].size(); ++ i) {
Edge &e = edges[G[x][i]];
if(!vis[e.to] && e.cap > e.flow) {
vis[e.to] = true;
dis[e.to] = dis[x] + 1;
que[++ tail] = e.to;
}
}
++ head;
}
return vis[t];
}
int dfs(int x, int a) {
if(x == t || a == 0) return a;
int flw = 0, f;
for(int &i = cur[x]; i < (signed) G[x].size(); ++ i) {
Edge &e = edges[G[x][i]];
if(dis[e.to] == dis[x] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
e.flow += f; edges[G[x][i] ^ 1].flow -= f; a -= f; flw += f;
if(a == 0) break;
}
}
return flw;
}
int maxflow(int s, int t) {
this->s = s; this->t = t;
int flw = 0;
while(bfs()) {
memset(cur, 0, sizeof cur);
flw += dfs(s, oo);
}
return flw;
}
void rebuild() {
for(int i = 0; i < (signed) edges.size(); ++ i)
edges[i].flow = 0;
}
void dfs(int u) {
mark[u] = true;
for(int i = 0; i < (signed) G[u].size(); ++ i) {
Edge e = edges[G[u][i]];
if(!mark[e.to] && e.cap > e.flow)
dfs(e.to);
}
}
}net;
void Divide(int l, int r) {
if(l >= r) return;
net.rebuild();
int nowflow = net.maxflow(a[l], a[r]);
memset(mark, false, (n + 1) * sizeof(bool));
net.dfs(a[l]);
for(int i = 1; i <= n; ++ i)
if(mark[i])
for(int j = 1; j <= n; ++ j)
if(!mark[j])
ans[i][j] = min(ans[i][j], nowflow), ans[j][i] = ans[i][j];
int L = l, R = r;
for(int i = l; i <= r; ++ i)
if(mark[a[i]]) tmp[L ++] = a[i];
else tmp[R --] = a[i];
for(int i = l; i <= r; ++ i)
a[i] = tmp[i];
Divide(l, L - 1); Divide(R + 1, r);
}
int main() {
int u, v, c;
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; ++ i) {
scanf("%d%d%d", &u, &v, &c);
net.add(u, v, c);
}
for(int i = 1; i <= n; ++ i) a[i] = i;
for(int i = 1; i <= n; ++ i)
for(int j = 1; j <= n; ++ j) ans[i][j] = oo;
net.n = n;
Divide(1, n);
for(int i = 1; i <= n; ++ i) {
for(int j = i + 1; j <= n; ++ j) {
lts.insert(ans[i][j]);
}
}
printf("%d\n", lts.size());
return 0;
}
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原文地址：http://www.cnblogs.com/sxprovence/p/5320043.html
